hdu 5600 N bulbs_bulbs csdn bfs-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_31186409/article/details/50411965

探讨了如何通过懒熊的特定路径改变灯泡状态，最终实现关闭所有灯的目标。文章提供了一个算法解决方案，用于判断是否能成功关闭所有灯，并给出示例。

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N bulbs

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 241 Accepted Submission(s): 132

Problem Description

N bulbs are in a row from left to right,some are on, and some are off.The first bulb is the most left one. And the last one is the most right one.they are numbered from 1 to n,from left to right.

in order to save electricity, you should turn off all the lights, but you're lazy.
coincidentally,a passing bear children paper(bear children paper means the naughty boy), who want to pass here from the first light bulb to the last one and leave.

he starts from the first light and just can get to the adjacent one at one step.
But after all,the bear children paper is just a bear children paper. after leaving a light bulb to the next one, he must touch the switch, which will change the status of the light.

your task is answer whether it's possible or not to finishing turning off all the lights, and make bear children paper also reach the last light bulb and then leave at the same time.

Input

The first line of the input file contains an integer T, which indicates the number of test cases.

For each test case, there are 2 lines.

The first line of each test case contains 1 integers n.

In the following line contains a 01 sequence, 0 means off and 1 means on.

*

1≤T≤10
*

1≤N≤1000000

Output

There should be exactly T lines in the output file.

The i-th line should only contain "YES" or "NO" to answer if it's possible to finish.

Sample Input

Sample Output

  
  
   
   YES


   
   
    
    
     
     Hint
    
    Child's path is: 123234545
all switchs are touched twice except the first one.

Source

BestCoder Round #67 (div.2)

好吧，依次把所有的数变成0，到最后一个如果是1就可以关闭所有的灯，否则就不可以

#include <iostream>
#include <cstdio>

using namespace std;

int a[1000000],n;

int dfs()
{
    for(int i = 0;i<n-1;i++)
    {
        if(!a[i])
        {
            a[i+1] = 1-a[i+1];
        }
    }
    return a[n-1];
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i<n;i++)
            scanf("%d",&a[i]);
        if(dfs())
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}